Question

Puck A of mass 240-g is traveling due east with a speed, v_Ai=10 m/s, on a level, frictionless air table when it collides with puck B of mass 160 g traveling at 40° south of west with a speed, v_Bi=15 m/s, on the same table. (See the diagram below.) When the pucks collide, they stick together via Velcro surfaces that line the circular boundaries of both pucks. Find the magnitude and direction of the momentum of the tandem of pucks after the collision. Draw the final momentum vector into the diagram below at the point of the collision.

Puck A of mass 240-g is traveling due east with a speed, vAi 10, on a level, frictionless air table when it collides with puck B of mass 160 g traveling at 40° south of west with a speed, VBi 15, on the same table. (See the diagram below.) When the pucks collide, they stick together via Velcro surfaces that line the circular boundaries of both pucks. Find the magnitude and direction of the momentum of the tandem of pucks after the collision. Draw the final momentum vector into the diagram below at the point of the collision. W+ E 40° VBI A B A VAI Collision

Answer 1

The concept involve in this question is the law of conservation of momentum before and after the collision , after collision puck A and puck B moves with the same velocity , The direction of the resultant momentum vector of the combined mass system of the puck A and puck B is same as the Resultant velocity of the puck A and puck B is being clearly shown by me .

Bi costo A B & NBisingoo Collision Momentum Before collision = MAVAitma Bi мл -Mass 4 че k A mot mass of fuck B Vai velocity of fuck A Vbi velocity of fuck B since After collision fuck stick togethen , so Momentum After collision = (ma+ma) W Z velocity of combined Mass system. Scanned W

Momentum Befor collision a Momentum After collision JAG = lo î . Voi = -15 80840 ? - 15 sin non MA= 240g , mp= 160g MATAP + M B u Bi = (MatuB) 340x loi + 160x1-1500846 în 15 sinho Å) = (240+160) 400 = 2400 î - 15x160 x 0.7660 î - 15X160X0.642 7 noot a 2400 î - 1838.4 î - 1540.8 Î 400u = 561.6 î - 1540.6 Î 1 3 = 1.404 î -3.852 Î -3.852 -17 L 1.404 a = tant { - 2.74358] d.oggle - 1 11.404)² + 13.8527 =

X 69.970 - Scanned with CamScanner

Momentum on the tandem of fucks After collisie = Imatra) = ( 240 + 16o) x 11.40 41 - 3.8527) - 561.6 î - 1540.8 Pm = 561.6 î - 1540.8 1 It is just the notation of the mome the combined mass system liml= (561.6²+(-1540.8² [1Pml = 1639.957 kgmis Qa dant 1-15408 561.6 To = -69.97° 69.9A Scanned with CamScanner

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